Newton polygon

In mathematics, the Newton polygon is a tool for understanding the behaviour of polynomials over local fields, or more generally, over ultrametric fields. In the original case, the local field of interest was essentially the field of formal Laurent series in the indeterminate X, i.e. the field of fractions of the formal power series ring $K[[X]]$, over $K$, where $K$ was the real number or complex number field. This is still of considerable utility with respect to Puiseux expansions. The Newton polygon is an effective device for understanding the leading terms $aX^{r}$ of the power series expansion solutions to equations $P(F(X))=0$ where $P$ is a polynomial with coefficients in $K[X]$, the polynomial ring; that is, implicitly defined algebraic functions. The exponents $r$ here are certain rational numbers, depending on the branch chosen; and the solutions themselves are power series in $K[[Y]]$ with $Y=X^{\frac {1}{d}}$ for a denominator $d$ corresponding to the branch. The Newton polygon gives an effective, algorithmic approach to calculating $d$.

After the introduction of the p-adic numbers, it was shown that the Newton polygon is just as useful in questions of ramification for local fields, and hence in algebraic number theory. Newton polygons have also been useful in the study of elliptic curves.

Definition

Construction of the Newton polygon of the polynomial 1 + 5 X + 1/5 X² + 35 X³ + 25 X⁵ + 625 X⁶ with respect to the 5-adic valuation.

A priori, given a polynomial over a field, the behaviour of the roots (assuming it has roots) will be unknown. Newton polygons provide one technique for the study of the behaviour of the roots.

Let $K$ be a field endowed with a non-archimedean valuation $v_{K}:K\to \mathbb {R} \cup \{\infty \}$, and let

$f(x)=a_{n}x^{n}+\cdots +a_{1}x+a_{0}\in K[x],$

with $a_{0}a_{n}\neq 0$. Then the Newton polygon of $f$ is defined to be the lower boundary of the convex hull of the set of points $P_{i}=\left(i,v_{K}(a_{i})\right),$ ignoring the points with $a_{i}=0$.

Restated geometrically, plot all of these points P_i on the xy-plane. Let's assume that the points indices increase from left to right (P₀ is the leftmost point, P_n is the rightmost point). Then, starting at P₀, draw a ray straight down parallel with the y-axis, and rotate this ray counter-clockwise until it hits the point P_k1 (not necessarily P₁). Break the ray here. Now draw a second ray from P_k1 straight down parallel with the y-axis, and rotate this ray counter-clockwise until it hits the point P_k2. Continue until the process reaches the point P_n; the resulting polygon (containing the points P₀, P_k1, P_k2, ..., P_km, P_n) is the Newton polygon.

Another, perhaps more intuitive way to view this process is this : consider a rubber band surrounding all the points P₀, ..., P_n. Stretch the band upwards, such that the band is stuck on its lower side by some of the points (the points act like nails, partially hammered into the xy plane). The vertices of the Newton polygon are exactly those points.

For a neat diagram of this see Ch6 §3 of "Local Fields" by JWS Cassels, LMS Student Texts 3, CUP 1986. It is on p99 of the 1986 paperback edition.

Main theorem

With the notations in the previous section, the main result concerning the Newton polygon is the following theorem, which states that the valuation of the roots of $f$ are entirely determined by its Newton polygon:

Let $\mu _{1},\mu _{2},\ldots ,\mu _{r}$ be the slopes of the line segments of the Newton polygon of $f(x)$ (as defined above) arranged in increasing order, and let $\lambda _{1},\lambda _{2},\ldots ,\lambda _{r}$ be the corresponding lengths of the line segments projected onto the x-axis (i.e. if we have a line segment stretching between the points $P_{i}$ and $P_{j}$ then the length is $j-i$).

• The $\mu _{i}$ are distinct;
• $\sum _{i}\lambda _{i}=n$;
• if $\alpha$ is a root of $f$ in $K$, $v(\alpha )\in \{-\mu _{1},\ldots ,-\mu _{r}\}$;
• for every $i$, the number of roots of $f$ whose valuations are equal to $-\mu _{i}$ (counting multiplicities) is at most $\lambda _{i}$, with equality if $f$ splits into the product of linear factors over $K$.

Corollaries and applications

With the notation of the previous sections, we denote, in what follows, by $L$ the splitting field of $f$ over $K$, and by $v_{L}$ an extension of $v_{K}$ to $L$.

Newton polygon theorem is often used to show the irreducibility of polynomials, as in the next corollary for example:

• Suppose that the valuation $v$ is discrete and normalized, and that the Newton polynomial of $f$ contains only one segment whose slope is $\mu$ and projection on the x-axis is $\lambda$. If $\mu =a/n$, with $a$ coprime to $n$, then $f$ is irreducible over $K$. In particular, since the Newton polygon of an Eisenstein polynomial consists of a single segment of slope $-{\frac {1}{n}}$ connecting $(0,1)$ and $(n,0)$, Eisenstein criterion follows.

Indeed, by the main theorem, if $\alpha$ is a root of $f$, $v_{L}(\alpha )=-a/n.$ If $f$ were not irreducible over $K$, then the degree $d$ of $\alpha$ would be $<n$, and there would hold $v_{L}(\alpha )\in {1 \over d}\mathbb {Z}$. But this is impossible since $v_{L}(\alpha )=-a/n$ with $a$ coprime to $n$.

Another simple corollary is the following:

• Assume that $(K,v_{K})$ is Henselian. If the Newton polygon of $f$ fulfills $\lambda _{i}=1$ for some $i$, then $f$ has a root in $K$.

Proof: By the main theorem, $f$ must have a single root $\alpha$ whose valuation is $v_{L}(\alpha )=-\mu _{i}.$ In particular, $\alpha$ is separable over $K$. If $\alpha$ does not belong to $K$, $\alpha$ has a distinct Galois conjugate $\alpha '$ over $K$, with $v_{L}(\alpha ')=v_{L}(\alpha )$, and $\alpha '$ is a root of $f$, a contradiction.

More generally, the following factorization theorem holds:

• Assume that $(K,v_{K})$ is Henselian. Then $f=A\,f_{1}\,f_{2}\cdots f_{r},$, where $A\in K$, $f_{i}\in K[X]$ is monic for every $i$, the roots of $f_{i}$ are of valuation $-\mu _{i}$, and $\deg(f_{i})=\lambda _{i}$.

Moreover, $\mu _{i}=v_{K}(f_{i}(0))/\lambda _{i}$, and if $v_{K}(f_{i}(0))$ is coprime to $\lambda _{i}$, $f_{i}$ is irreducible over $K$.

Proof: For every $i$, denote by $f_{i}$ the product of the monomials $(X-\alpha )$ such that $\alpha$ is a root of $f$ and $v_{L}(\alpha )=-\mu _{i}$. We also denote $f=AP_{1}^{k_{1}}P_{2}^{k_{2}}\cdots P_{s}^{k_{s}}$ the factorization of $f$ in $K[X]$ into prime monic factors $(A\in K).$ Let $\alpha$ be a root of $f_{i}$. We can assume that $P_{1}$ is the minimal polynomial of $\alpha$ over $K$. If $\alpha '$ is a root of $P_{1}$, there exists a K-automorphism $\sigma$ of $L$ that sends $\alpha$ to $\alpha '$, and we have $v_{L}(\sigma \alpha )=v_{L}(\alpha )$ since $K$ is Henselian. Therefore $\alpha '$ is also a root of $f_{i}$. Moreover, every root of $P_{1}$ of multiplicity $\nu$ is clearly a root of $f_{i}$ of multiplicity $k_{1}\nu$, since repeated roots share obviously the same valuation. This shows that $P_{1}^{k_{1}}$ divides $f_{i}.$ Let $g_{i}=f_{i}/P_{1}^{k_{1}}$. Choose a root $\beta$ of $g_{i}$. Notice that the roots of $g_{i}$ are distinct from the roots of $P_{1}$. Repeat the previous argument with the minimal polynomial of $\beta$ over $K$, assumed w.l.g. to be $P_{2}$, to show that $P_{2}^{k_{2}}$ divides $g_{i}$. Continuing this process until all the roots of $f_{i}$ are exhausted, one eventually arrives to $f_{i}=P_{1}^{k_{1}}\cdots P_{m}^{k_{m}}$, with $m\leq s$. This shows that $f_{i}\in K[X]$, $f_{i}$ monic. But the $f_{i}$ are coprime since their roots have distinct valuations. Hence clearly $f=Af_{1}\cdot f_{2}\cdots f_{r}$, showing the main contention. The fact that $\lambda _{i}=\deg(f_{i})$ follows from the main theorem, and so does the fact that $\mu _{i}=v_{K}(f_{i}(0))/\lambda _{i}$, by remarking that the Newton polygon of $f_{i}$ can have only one segment joining $(0,v_{K}(f_{i}(0))$ to $(\lambda _{i},0=v_{K}(1))$. The condition for the irreducibility of $f_{i}$ follows from the corollary above. (q.e.d.)

The following is an immediate corollary of the factorization above, and constitutes a test for the reducibility of polynomials over Henselian fields:

• Assume that $(K,v_{K})$ is Henselian. If the Newton polygon does not reduce to a single segment $(\mu ,\lambda ),$ then $f$ is reducible over $K$.

Other applications of the Newton polygon comes from the fact that a Newton Polygon is sometimes a special case of a Newton polytope, and can be used to construct asymptotic solutions of two-variable polynomial equations like $3x^{2}y^{3}-xy^{2}+2x^{2}y^{2}-x^{3}y=0.$

This diagram shows the Newton polygon for P(x,y) = 3x² y³ − xy² + 2x²y² − x³y, with positive monomials in red and negative monomials in cyan. Faces are labelled with the limiting terms they correspond to.

Symmetric function explanation

In the context of a valuation, we are given certain information in the form of the valuations of elementary symmetric functions of the roots of a polynomial, and require information on the valuations of the actual roots, in an algebraic closure. This has aspects both of ramification theory and singularity theory. The valid inferences possible are to the valuations of power sums, by means of Newton's identities.

History

Newton polygons are named after Isaac Newton, who first described them and some of their uses in correspondence from the year 1676 addressed to Henry Oldenburg.

See also

• F-crystal
• Eisenstein's criterion
• Newton–Okounkov body
• Newton polytope

References

• Goss, David (1996), Basic structures of function field arithmetic, Ergebnisse der Mathematik und ihrer Grenzgebiete (3) [Results in Mathematics and Related Areas (3)], vol. 35, Berlin, New York: Springer-Verlag, doi:10.1007/978-3-642-61480-4, ISBN 978-3-540-61087-8, MR 1423131
• Gouvêa, Fernando: p-adic numbers: An introduction. Springer Verlag 1993. p. 199.

External links

• Applet drawing a Newton Polygon